I am try to determine
What is the degree of the splitting field of $x^5 − 7$ over $\mathbb{Q}$?
I concluded the degree of $\mathbb{Q}[\sqrt[5]{7}]$ over $\mathbb{Q}$ is 5 because the polynomial is irreducible in $\mathbb{Q}$. And that the solutions to the polynomial are $\alpha = \sqrt[5]{7}$ times the 5th roots of unity $\beta = e^{2\pi i/5}$, that is $\alpha$, $\alpha\beta$, $\alpha\beta^2$, $\alpha\beta^3$, and $\alpha\beta^4$. Furthermore the degree of $\mathbb{Q}[\beta]$ over $\mathbb{Q}$ is 4. Thus the degree of the splitting field is $[\mathbb{Q}[\alpha,\beta]:\mathbb{Q}]$ = 20, because 4 and 5 have no common divisors.
Is this correct?
$\newcommand{\rt}{\sqrt[5]7}$ $\newcommand{\Q}{\mathbb{Q}}$
Yes, you are right.
Let $\zeta := \exp(2 \pi i / 5)$, then the splitting field of your polynomial is $\mathbb{Q}(\rt, \zeta \rt, \dots, \zeta^4\rt) = \Q(\zeta, \rt)$. The degree of $\Q(\zeta)/\Q = 4$, because either you know that $Gal(\Q(\zeta)/\Q) \cong (\mathbb{Z}/5 \mathbb{Z})^x$ (where $\mathbb Z / 5 \mathbb Z$ is a field, thus the only non-unit is $0$) or you can also easily show that $\zeta$ is a root of $\Phi_5(X) = X^4+X^3+X^2+X+1$ and that this polynomial is irreducible.
$[\Q(\rt): \Q] = 5$, as your given polynomial is irreducible (Eisenstein).
Then, as you have already mentioned, the degree of the big extension must be $20$ as $(4,5) = 1$